Question: $\dfrac{d}{dx}\left(-\dfrac{1}{x^5}-\dfrac{1}{x^3}+\dfrac{1}{x}\right)=$
Answer: The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}-\dfrac{1}{x^5}-\dfrac{1}{x^3}+\dfrac{1}{x} \\\\ &=-x^{-5}-x^{-3}+x^{-1} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-x^{-5}-x^{-3}+x^{-1}) \\\\ &=-1\dfrac{d}{dx}(x^{-5})-\dfrac{d}{dx}(x^{-3})+\dfrac{d}{dx}(x^{-1}) \\\\ &=-1(-5x^{-6})-(-3)x^{-4}+(-1)x^{-2} \\\\ &=5x^{-6}+3x^{-4}-1x^{-2} \\\\ &=\dfrac{5}{x^6}+\dfrac{3}{x^4}-\dfrac{1}{x^2} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(-\dfrac{1}{x^5}-\dfrac{1}{x^3}+\dfrac{1}{x}\right)=\dfrac{5}{x^6}+\dfrac{3}{x^4}-\dfrac{1}{x^2}$.